Not using Jacobi symbol how to prove For all positive integer $n>1$ $2^n - 1 \not | 3^n-1$?

85 Views Asked by At

There is a proof:

if $n$ is even,then $3|2^n-1$ but $3\not|\;3^n-1$,It is correct;

if $n$ is odd,suppose $2^n-1|3^n-1$,then $3^n \equiv 1(\mod 2^n-1)$,then $$\left(\frac{3^n}{2^n-1}\right)=\left(\frac{1}{2^n-1}\right)=1.$$ However $$\left(\frac{3^n}{2^n-1}\right)=\left(\frac{3}{2^n-1}\right)^n=\left(\frac{3}{2^n-1}\right)=-\left(\frac{2^n-1}{3}\right)=-\left(\frac{1}{3}\right)=-1$$ That is a contradiction.

$\left(\frac{m}{n}\right)$ is Jacobi symbol,which is a generalization of Legendre symbol.

I want a proof without high technique. So I transformed the right side for $$3^n-1=(2+1)^n-1=2(2^{n-1}+n\times2^{n-2}+\cdots n),$$ and then have no idea. Can this way work?